# How do you find (d^2y)/(dx^2) given x^2+xy-y^2=1?

Oct 28, 2016

$y ' ' \left(x\right) = \pm \frac{10}{5 {x}^{2} - 4} ^ \left(\frac{3}{2}\right)$

#### Explanation:

Calling

$f \left(x , y \left(x\right)\right) = {x}^{2} + x y \left(x\right) - y {\left(x\right)}^{2} - 1 = 0$

$\frac{\mathrm{df}}{\mathrm{dx}} = 2 x + y \left(x\right) - 2 y \left(x\right) y ' \left(x\right) = 0$

solving for $y ' \left(x\right)$

$y ' \left(x\right) = - \frac{2 x + y \left(x\right)}{x - 2 y \left(x\right)}$

computing now

d/dx((df)/dx) =2 + 2 y'(x) - 2 y'(x)^2 + x y''(x) - 2 y(x) y''(x)=0  solving for $y ' ' \left(x\right)$

$y ' ' \left(x\right) = \frac{10 \left({x}^{2} + x y \left(x\right) - y {\left(x\right)}^{2}\right)}{x - 2 y \left(x\right)} ^ 3$ after substituting $y ' \left(x\right)$

Finally substituting $y \left(x\right) = \frac{1}{2} \left(x \pm \sqrt{5 {x}^{2} - 4}\right)$ we obtain

$y ' ' \left(x\right) = \pm \frac{10}{5 {x}^{2} - 4} ^ \left(\frac{3}{2}\right)$